Answer:
moles of ammonia produced = 0.28 moles
Explanation:
The reaction is

As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia
So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen
= 0.42 moles of hydrogen molecule.
this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia
the moles of ammonia produced = 0.28 moles
Here the nitrogen is limiting reagent.
N2 + 3H2 ----> 2NH3
<span>you can see 3 moles H2 reacts to form 2 moles NH3 </span>
<span>Therefore moles NH3 = 2 / 3 x moles H2 </span>
<span>= 2/3 x 12.0 mol </span>
<span>= 8.00 mol NH3 hope this help</span>
Answer: 
Explanation:

cM 0 0
So dissociation constant will be:

Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)


As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)

![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of
for the acid is 
Answer:
35.9%
Explanation:
The percent volume of the coffee solution can be calculated as follows:
% volume of coffee solution = volume of coffee/total volume of coffee solution × 100
According to this question, a cup of coffee has 71 mL of coffee and 127 mL of water. This means that, the total volume of coffee solution is;
71mL + 127mL = 198mL
% volume = 71/198 × 100
= 0.359 × 100
Percent volume of coffee solution = 35.9%